I can't think of a good title for this

This is a special case of formula 12 here ; the proof of a much more general result can be found here , formulas 30 and 31 . For $s=2$ it's $\Gamma(4)\zeta(4)=3!\times \frac{\pi^4}{90}=\frac{\pi^4}{15}$ , so that the answer is $\boxed{20}$

First Let us make the substitution $xy =t$

Then the integral becomes:-

$\Large \int_{0}^{1}\int_{0}^{y}\frac{\ln^{2}(t)}{y(1-t)} dt dy$

Now changing the order of integration we have:-

$\Large \int_{0}^{1}\int_{t}^{1}\frac{\ln^{2}(t)}{y(1-t)} dy dt$

We have :-

$\Large -\int_{0}^{1}\frac{\ln^{3}(t)}{(1-t)}dt$

We would now use the infinite geometric sum formula

$\displaystyle \frac{1}{1-x} = \sum_{r=0}^{\infty} x^{r}$

So we have our integral as :-

$\Large -\sum_{r=0}^{\infty}\int_{0}^{1}\ln^{3}(t)t^{r} dt$

Now using the substitution $t=e^{-z}$ we have:-

$\Large \sum_{r=0}^{\infty}\int_{0}^{\infty}z^{3}e^{-(r+1)z} dz$

Now again substituting $(r+1)z = x$ we have:-

$\Large \sum_{r=0}^{\infty}\int_{0}^{\infty}\frac{x^{3}e^{-x}}{(r+1)^{4}} dx$

$\Large =\sum_{r=0}^{\infty}\frac{\Gamma(4)}{(r+1)^{4}}=\Gamma(4)\zeta(4) = \frac{\pi^{4}}{15}$

Here $\Gamma(.)$ denotes the Gamma Function and $\zeta(s)$ denotes the Riemann Zeta Function